Solve (3y+1)(3y-1)-left(3y-4right)^2=7

$$(3y+1)(3y-1)- { \left(3y-4 \right) }^{ 2 } =7$$

Solve for y

$$y=1$$

Steps for Solving Linear Equation

Consider \left(3y+1\right)\left(3y-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3y\right)^{2}-1-\left(3y-4\right)^{2}=7$$

Expand \left(3y\right)^{2}.

$$3^{2}y^{2}-1-\left(3y-4\right)^{2}=7$$

Calculate 3 to the power of 2 and get 9.

$$9y^{2}-1-\left(3y-4\right)^{2}=7$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-4\right)^{2}.

$$9y^{2}-1-\left(9y^{2}-24y+16\right)=7$$

To find the opposite of 9y^{2}-24y+16, find the opposite of each term.

$$9y^{2}-1-9y^{2}+24y-16=7$$

Combine 9y^{2} and -9y^{2} to get 0.

$$-1+24y-16=7$$

Subtract 16 from -1 to get -17.

$$-17+24y=7$$

Add 17 to both sides.

$$24y=7+17$$

Add 7 and 17 to get 24.

$$24y=24$$

Divide both sides by 24.

$$y=\frac{24}{24}$$

Divide 24 by 24 to get 1.

$$y=1$$